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求这道不定积分的解题步骤
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∫√(1+sinx)dx
=2∫√[sin(x/2)+cos(x/2)]²d(x/2)
=2∫[sin(x/2)+cos(ⅹ/2)]d(x/2)
=sin(x/2)-cos(x/2)+C
=2∫√[sin(x/2)+cos(x/2)]²d(x/2)
=2∫[sin(x/2)+cos(ⅹ/2)]d(x/2)
=sin(x/2)-cos(x/2)+C
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第一步没看懂
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